0=(-16t^2)-96t+160

Solution for 0=(-16t^2)-96t+160 equation:

0=(-16t^2)-96t+160

We move all terms to the left:

0-((-16t^2)-96t+160)=0

We add all the numbers together, and all the variables

-((-16t^2)-96t+160)=0


We calculate terms in parentheses: -((-16t^2)-96t+160), so:

(-16t^2)-96t+160

We get rid of parentheses

-16t^2-96t+160

Back to the equation:

-(-16t^2-96t+160)


We get rid of parentheses

16t^2+96t-160=0

a = 16; b = 96; c = -160;
Δ = b2-4ac
Δ = 962-4·16·(-160)
Δ = 19456
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{19}}{2*16}=\frac{-96-32\sqrt{19}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{19}}{2*16}=\frac{-96+32\sqrt{19}}{32}
$